| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\tan(75^\circ) = \frac{v_y}{v_x}\) | To find the angle of the velocity vector, we use the relationship between the vertical and horizontal components. Here, \(v_y\) is the vertical speed, and \(v_x = 2.7 \, \text{m/s}\) is the horizontal speed, and we are given the angle is \(75^\circ\). |
| 2 | \( \tan(75^\circ) = \frac{v_y}{2.7}\) | Substitute the given horizontal speed \(v_x = 2.7 \, \text{m/s}\) into the equation. |
| 3 | \(v_y = 2.7 \tan(75^\circ)\) | Solve for the vertical speed \(v_y\). |
| 4 | \(\tan(75^\circ) \approx 3.73\) | Use a calculator to find the value of \(\tan(75^\circ)\). |
| 5 | \(v_y = 2.7 \times 3.73\) | Substitute the known value of \(\tan(75^\circ)\). |
| 6 | \(v_y \approx 10.07 \, \text{m/s}\) | Multiply to find the vertical speed \(v_y\). |
| 7 | \(v_y = v_i + at\) | Use the kinematic equation to solve for the time \(t\). Initial vertical velocity \(v_i = 0 \), acceleration \(a = g\) (where \(g = 9.8 \, \text{m/s}^2\)), and final vertical velocity \(v_y \approx 10.07 \, \text{m/s}\). |
| 8 | \(10.07 = 0 + 9.8 t\) | Substitute the known values into the kinematic equation. |
| 9 | \(t = \frac{10.07}{9.8}\) | Isolate the time \(t\). |
| 10 | \(t \approx 1.03 \, \text{s}\) | Solve for \(t\). |
| 11 | \Delta x = v_i t + \frac{1}{2} a t^2\) | Use the vertical displacement formula to find \(\Delta x\). Here, \(v_i = 0\), \(a = g\), and we need to find \(\Delta x\) for the time \(t\). |
| 12 | \( \Delta x = 0 + \frac{1}{2} (9.8)(1.03)^2\) | Plug in the values for \(a\) and \(t\). |
| 13 | \(\Delta x \approx \frac{1}{2} (9.8)(1.0609)\) | Simplify inside the parentheses. |
| 14 | \(\Delta x \approx 5.20 \, \text{m}\) | Calculate the final displacement. |
| 15 | \boxed{\Delta x \approx 5.20 \, \text{m}}\) | Final vertical distance below the edge where the velocity vector points downward at a \( 75^\circ \) angle. |
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A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island? Use \( g = 9.81 \, m/s^2\).
A rocket-powered hockey puck has a thrust of \(4.40 \, \text{N}\) and a total mass of \(1.00 \, \text{kg}\). It is released from rest on a frictionless table, \(2.10 \, \text{m}\) from the edge of a \(2.10 \, \text{m}\) drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket is present for the entire time of travel, how far does the puck land from the base of the table?
A ball is launched and lands \( 20 \) \( \text{m} \) away below the launch point \( 2.5 \) \( \text{s} \) later. The maximum height reached is \( 8.0 \) \( \text{m} \). What was the original launch velocity?
A officer fires a pistol horizontally toward a target \(120 \,\text{m}\) at a velocity of \(200 \, \text{m/s}\). If the officer aimed directly at the bull’s eye
The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
A cannon fires projectiles on a flat range at a fixed speed but with variable angle. The maximum range of the cannon is \(L\). What is the range of the cannon when it fires at an angle of \(30^\circ\) above the horizontal? Ignore air resistance.
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of \( 150 \) \( \text{km/h} \), by what \( \% \) will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, \( 18 \) \( \text{m} \) away? For this estimate, ignore air resistance and spin on the ball.
Projectiles 1 and 2 are launched from level ground at the same time and follow the trajectories shown in the figure. Which one of the projectiles, if either, returns to the ground first, and why?
5.2 m
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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